- --- On Thu, 11/8/12, Ludovicus <luiroto@...> wrote:
> "I we decompose any prime p such that

Yup:

> p = x + y, there are (p-1)/2 forms of doing that. Of those,

> many can produce: x^2 + y^2 = prime.

>

> CONJECTURE. Calling #prime2 that quantity of primes, we have:

> Be R = #prime2*Log(p) / p . Then Lim. R = 0.70..

> when p--->inf.

>

> Example. p = 40037 ; #prime2 = 2641 ; R = 0.6991

? lcv(p)=sum(x=1,(p-1)/2,ispseudoprime(x^2+(p-x)^2))

? lcv(40037)

2641

> From p = 200003 to 100271 the mean R = 0.700205.

Note that a mean tells you nothing about a limit, in fact, the mean may hide the fact that there's no limit, if it fluctuates between a liminf and a limsup. However, the mean is an equally valid value to be interested in, so that's not a problem.

> I will call that conjecture : "Conjecture Ludovicus V"

Not enough data to deny it, and some anecdotal evidence to support it:

? lcv(p)=sum(x=1,(p-1)/2,ispseudoprime(x^2+(p-x)^2))*log(p)/p

time = 0 ms.

? for(n=1,30,p=nextprime(2^n);print(n" "p" "lcv(p)))

1 2 0

2 5 0.6437751649736401498403037333

3 11 0.6539714380359192392896209758

4 17 0.6666384338955802541763610866

5 37 0.7807390081392917717552639289

6 67 0.6903226688552332336771759994

7 131 0.8187354283238575112312009790

8 257 0.7341190151223247982255292090

9 521 0.7204318666126718139833661082

10 1031 0.6258588331848470155706204411

11 2053 0.6910047148395137088493983438

12 4099 0.7082595544903668720682217777

13 8209 0.6982896023307951368923678478

14 16411 0.6978693804400682993622951493

15 32771 0.7097359994440564892776086479

16 65537 0.7004141483596733103866463592

17 131101 0.6948838285968673741945871092

18 262147 0.6944460786798661063812187741

19 524309 0.6969871254592594758140428033

20 1048583 0.6992402515337114838314592173

21 2097169 0.7007186928388776062866463444

22 4194319 0.6970925620012833876499535047

23 8388617 0.6951323183116127789119946883

24 16777259 0.6948711428730801610212743660

25 33554467 0.6956608736980247707005160260

^C

This is probably not the correct way to approach the problem, as it only produces.

A better way might be to split prime p=4n+1 into x^2+y^2 (Cornaccia-Smith, or simply factor in Z[I]), tally x+y, and then for each prime q, once sqrt(2p) exceeds q, read off the value of the tally for q.

This script is a starting point, but doesn't care about the q values, just makes the tallies:

? lim=1000;tally=vectorsmall(lim/2);forprime(p=5,lim*lim/2,if(p%4==1,f=factor(p+0*I)[2,1];p2=real(f)+imag(f);tally[(p2+1)/2]++));tally

time = 626 ms.

Vecsmall([0, 1, 2, 2, 2, 3, 3, 4, 4, 4, 3, 4, 8, 4, 6, 5, 4, 9, 8, 6, 9, 7, 7, 7, 5, 7, 9, 14, 8, 9, 11, 7, 17, 11, 10, 9, 11, 9, 8, 13, 9, 15, 20, 11, 14, 13, 8, 18, 14, 10, 18, 16, 10, 17, 16, 13, 20, 20, 13, 14, 17, 12, 23, 18, 14, 22, 15, 17, 18, 21, 12, 19, 29, 16, 23, 21, 14, 27, 24, 14, 19, 24, 19, 22, 25, 16, 24, 24, 13, 22, 28, 14, 35, 25, 15, 30, 23, 26, 27, 24, 17, 26, 36, 18, 21, 29, 22, 39, 27, 23, 32, 30, 27, 30, 28, 17, 29, 43, 20, 24, 28, 17, 38, 34, 20, 30, 28, 31, 34, 27, 23, 33, 48, 24, 36, 35, 26, 41, 31, 23, 37, 32, 33, 27, 38, 22, 36, 53, 20, 38, 35, 26, 53, 37, 28, 41, 42, 33, 42, 37, 28, 44, 58, 25, 35, 42, 25, 58, 43, 22, 37, 37, 36, 42, 38, 29, 34, 57, 23, 43, 39, 26, 60, 38, 31, 42, 45, 49, 56, 44, 32, 44, 45, 28, 44, 42, 32, 58, 42, 20, 45, 52, 42, 48, 51, 33, 48, 68, 28, 47, 50, 35, 76, 43, 38, 48, 49, 46, 51, 46, 27, 48, 69, 39, 57, 54, 34, 62, 57, 44, 53, 53, 49, 53, 42, 39, 47, 75, 38, 60, 59, 25, 81, 57, 31, 52, 65, 39,

47, 59, 32, 51, 74, 42, 54, 48, 30, 76, 52, 41, 60, 56, 51, 62, 62, 45, 67, 82, 41, 48, 69, 40, 82, 58, 47, 63, 56, 52, 57, 77, 42, 68, 90, 34, 61, 67, 43, 80, 67, 43, 52, 56, 66, 67, 57, 40, 70, 77, 40, 68, 62, 50, 75, 68, 35, 77, 71, 60, 66, 67, 45, 55, 90, 38, 77, 69, 48, 90, 69, 45, 68, 68, 62, 74, 66, 32, 72, 88, 49, 72, 67, 64, 82, 73, 51, 66, 65, 62, 76, 63, 49, 76, 98, 55, 86, 74, 38, 105, 81, 48, 70, 77, 72, 61, 83, 56, 69, 94, 54, 74, 56, 51, 110, 75, 51, 72, 72, 53, 67, 70, 55, 84, 103, 45, 59, 73, 46, 88, 72, 33, 68, 61, 61, 79, 66, 53, 71, 92, 39, 67, 69, 42, 84, 63, 38, 48, 72, 64, 70, 67, 42, 61, 73, 44, 61, 67, 42, 79, 54, 39, 63, 63, 46, 61, 62, 44, 59, 73, 39, 61, 60, 35, 78, 41, 35, 55, 53, 45, 54, 51, 31, 43, 66, 40, 47, 59, 41, 65, 50, 39, 54, 46, 43, 45, 39, 31, 42, 62, 35, 41, 50, 26, 56, 46, 29, 45, 34, 40, 35, 38, 32, 42, 48, 23, 45, 29, 28, 50, 39, 21, 37, 30, 27, 29, 40, 24, 29, 35, 21, 25, 29, 18, 42, 30, 16, 22, 23, 22, 20,

20, 14, 19, 21, 12, 15, 14, 10, 14, 10, 2])

YMMV...

Phil - --- In primenumbers@yahoogroups.com,

Phil Carmody <thefatphil@...> wrote:

> 17 131101 0.6948838285968673741945871092

Looks like an expensive method of approximating log(2) ?

> 18 262147 0.6944460786798661063812187741

> 19 524309 0.6969871254592594758140428033

> 20 1048583 0.6992402515337114838314592173

> 21 2097169 0.7007186928388776062866463444

> 22 4194319 0.6970925620012833876499535047

> 23 8388617 0.6951323183116127789119946883

> 24 16777259 0.6948711428730801610212743660

> 25 33554467 0.6956608736980247707005160260

David (with no sound basis for this guess) - Le 2012-11-11 13:11, djbroadhurst a écrit :
>> 25 33554467 0.6956608736980247707005160260

hmmmm you mean ln(2) :-)

> Looks like an expensive method of approximating log(2) ?

> David (with no sound basis for this guess)

Yann (who wonders what kind of relationship

there can be between ln(2) and the threshold

voltage of a bipolar transistor...

damn coincidences) - --- In primenumbers@yahoogroups.com,

whygee@... asked:

> hmmmm you mean ln(2) :-)

ln(2) is just a fussy way of writing log(2) for the

benefit people who think that God has 10 fingers.

In fact She has exp(1) fingers :-)

David - --- On Sun, 11/11/12, djbroadhurst <d.broadhurst@...> wrote:
> --- In primenumbers@yahoogroups.com, Phil Carmody <thefatphil@...> wrote:

That value went through my head too, certainly. If it is log(2), I would hope that there is a sound reason for it to be so.

> > 17 131101 0.6948838285968673741945871092

> > 18 262147 0.6944460786798661063812187741

> > 19 524309 0.6969871254592594758140428033

> > 20 1048583 0.6992402515337114838314592173

> > 21 2097169 0.7007186928388776062866463444

> > 22 4194319 0.6970925620012833876499535047

> > 23 8388617 0.6951323183116127789119946883

> > 24 16777259 0.6948711428730801610212743660

> > 25 33554467 0.6956608736980247707005160260

>

> Looks like an expensive method of approximating log(2) ?

>

> David (with no sound basis for this guess)

Phil